402 11 Differential calculus on manifolds
Exercise 11.11: Use the infinitesimal homotopy relation, to show that L
v
H
ω = 0, where
v
H
is the vector field corresponding to H . Suppose now that the phase space is 2n
dimensional. Show that in local Darboux coordinates the 2n-form ω
n
/n! is, up to a sign,
the phase-space volume element d
n
pd
n
q. Show that L
v
H
ω
n
/n!=0 and that this result
is Liouville’s theorem on the conservation of phase-space volume.
The classical mechanics of spin
It is sometimes said in books on quantum mechanics that the spin of an electron, or other
elementary particle, is a purely quantum concept and cannot be described by classical
mechanics. This statement is false, but spin is the simplest system in which traditional
physicists’ methods become ugly and it helps to use the modern symplectic language. A
“spin” S can be regarded as a fixed length vector that can point in any direction in R
3
.
We will take it to be of unit length so that its components are
S
x
= sin θ cos φ,
S
y
= sin θ sin φ,
S
z
= cos θ, (11.106)
where θ and φ are polar coordinates on the 2-sphere S
2
.
The surface of the sphere turns out to be both the configuration space and the phase
space. In particular the phase space for a spin is not the cotangent bundle of the configu-
ration space. This has to be so: we learned from Niels Bohr that a 2n-dimensional phase
space contains roughly one quantum state for every
n
of phase-space volume. A cotan-
gent bundle always has infinite volume, so its corresponding Hilbert space is necessarily
infinite dimensional. A quantum spin, however, has a finite-dimensional Hilbert space
so its classical phase space must have a finite total volume. This finite-volume phase
space seems unnatural in the traditional view of mechanics, but it fits comfortably into
the modern symplectic picture.
We want to treat all points on the sphere alike, and so it is natural to take the symplectic
2-form to be proportional to the element of area. Suppose that ω = sin θ dθdφ. We could
write ω =−d cos θ dφ and regard φ as “q” and −cos θ as “p” (Darboux’ theorem in
action!), but this identification is singular at the north and south poles of the sphere, and,
besides, it obscures the spherical symmetry of the problem, which is manifest when we
think of ω as d(area).
Let us take our Hamiltonian to be H = BS
x
, corresponding to an applied magnetic
field in the x-direction, and see what Hamilton’s equations give for the motion. First we
take the exterior derivative
d(BS
x
) = B(cos θ cos φdθ − sin θ sin φdφ). (11.107)
This is to be set equal to
−ω(v
BS
x
, ) = v
θ
(−sin θ)dφ +v
φ
sin θdθ . (11.108)